Question

Solve the equation $\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$ where $a+b+c\ne 0$.

• A. $x=a+b+c,\pm \sqrt{\frac{1}{2}\{{(a-b)}^2-{(b-c)}^2-{(c-a)}^2\}}$
• B. $x=a+b+c,\pm \sqrt{\frac{1}{2}\{{(a-b)}^2+{(b-c)}^2+{(c-a)}^2\}}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $x=a+b+c,\pm \sqrt{\frac{1}{2}\{{(a-b)}^2+{(b-c)}^2+{(c-a)}^2\}}$

$\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$

$C_1\to C_1+C_2+C_3$

$\left|\begin{array}{ccc}a+b+c-x& c& b\\ a+b+c-x& b-x& a\\ a+b+c-x& a& c-x\end{array}\right|=0$

Taking $a+b+c-x$ common from first column,

$(a+b+c-x)\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|=0$

Now $R_2\to R_2-R_1,R_3\to R_3-R_1$

$(a+b+c-x)\left|\begin{array}{ccc}1& c& b\\ 0& b-c-x& a-b\\ 0& a-c& c-b-x\end{array}\right|=0$

Expanding along first column and simplifying we get,

$(a+b+c)(x^2-\frac{1}{2}\{{(a-b)}^2+{(b-c)}^2+{(c-a)}^2\})=0$

$\therefore x=a+b+c,\pm \sqrt{\frac{1}{2}\{{(a-b)}^2+{(b-c)}^2+{(c-a)}^2\}}$